Ask question

Ask question

All categories

3 years ago

10

is there difference if you pull a block down a hill or if you push a block down a hill. Or if you push a block up a hill or pull

it up ? For example if you PULL a block down a hill it would be F+Fg||-fk =ma (the force you are using +Fsinθ - kinetic friction =ma. Would this change if you PUSH the block down?

1 answer:

Afina-wow [57]3 years ago

8 0

Answer:

Yes it will be changed

Explanation:

Lifting up requires potential energy while dropping down requires kinetic, both will have different equations

You might be interested in

What would you need to do to find the number of neutrons in an atom?

I am Lyosha [343]

Answer:

You subtract the atomic number from the atomic mass to find the number of neutrons alone

Explanation:

The atomic number is the number of protons OR electrons. They are both equal thus making the atom have a neutral charge. The relative atomic mass is the number of protons AND neutrons.

3 0

3 years ago

Grace is given a piece of wood, an iron nail, and a styrofoam cup. She puts these objects into a bucket of water and observes th

wel

Density because any object that is more dense than water (density of 1. I forget the units) sinks, and any object less dense than water floats.

7 0

3 years ago

A 2120 kg car traveling at 13.4 m/s collides with a 2810 kg car that is initally at rest at a stoplight. The cars stick together

viktelen [127]

Answer:

The coefficient of friction between the cars and the road is 0.859.

Explanation:

The two cars collide each other inelastically, then we can determine the resulting velocity by the Principle of Momentum Conservation:

$m_{A}\cdot v_{A} + m_{B}\cdot v_{B} = (m_{A} + m_{B})\cdot v$ (1)

Where:

$m_{A}$ , $m_{B}$ - Masses of the cars, in kilograms.

$v_{A}$ , $v_{B}$ - Initial velocities of the cars, in meters per second.

$v$ - Velocity of the resulting system, in meters per second.

If we know that $m_{A} = 2120\,kg$ , $v_{A} = 13.4\,\frac{m}{s }$ , $m_{B} = 2810\,kg$ and $v_{B} = 0\,\frac{m}{s}$ , then the velocity of the resulting system:

$v = \frac{m_{A}\cdot v_{A}+m_{B}\cdot v_{B}}{m_{A}+m_{B}}$

$v = \frac{(2120\,kg)\cdot \left(13.4\,\frac{m}{s} \right)+(2810\,kg)\cdot \left(0\,\frac{m}{s} \right)}{2120\,kg + 2810\,kg}$

$v = 5.762\,\frac{m}{s}$

By Principle of Energy Conservation and Work-Energy Theorem, we understand that the initial translational kinetic energy ( $K$ ), in joules, is dissipated due to work done by friction ( $W_{f}$ ), in joules, that is to say:

$K = W_{f}$ (2)

$\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2} = \mu\cdot (m_{A}+m_{B})\cdot g \cdot s$

$\frac{1}{2}\cdot v^{2} = \mu \cdot g\cdot s$ (2b)

Where:

$\mu$ - Coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Travelled distance, in meters.

If we know that $v = 5.762\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $s = 1.97\,m$ , then the coefficient of friction is:

$\mu = \frac{v^{2}}{2\cdot g\cdot s}$

$\mu = \frac{\left(5.762\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.97\,m)}$

$\mu = 0.859$

The coefficient of friction between the cars and the road is 0.859.

6 0

3 years ago

The flywheel in the shape of a solid cylinder of mass 75.0 kg and diameter 1.40 m rotates about an axis passing through the cent

Anni [7]

The torque required for the cylinder is 76.99 N m.

What is torque?

Torque may be defined as the product of the force and the distance perpendicular to the force from the axis of rotation.

Calculation of the angular acceleration:

The first kinematic equation for the rotational motion is,

ω=ω₀+αt

where ω₀ is the initial angular velocity, ω is the final angular velocity, α is the angular acceleration and t is the time. Since the cylinder starts from the rest, its initial angular velocity is zero.

Here, ω=320 rev/min or ω=320*2π/60 rad/sec, ω₀=0 and t=8.00 sec. Put these values in the above equation and solve it to get the value of angular acceleration.

320*2π/60=0+α*8

α=320*2π/(8*60)

α=4.19 rad/s^2

Calculation of the torque:

The torque T is given by the formula,

T=Iα

where I is the moment of inertia.

For the cylinder, I=1/2*mr^2

where m is the mass of the cylinder and r is the radius of the cylinder.

So,

T=1/2*mr^2*α

Here m=75 kg, radius r is half of the diameter r=1.4/2=0.7 m, and α=4.19 rad/s^2. Using these values in the formula of torque.

T=1/2*(75)*(0.7)^2*4.19

T= 76.99 N m

Learn more about the torque.

brainly.com/question/16049994

#SPJ4

6 0

1 year ago

Which of the following is not a measurement?

navik [9.2K]

Answer:

120 is the correct answer

4 0

2 years ago